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 [2014] tommy's theorem sexp ' (z) =/= 0 ? tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 06/17/2014, 12:18 PM Let sexp(z) be a solution that is analytic in the entire complex plane apart from z=-2,-3,-4,... if w is a (finite) nonreal complex number such that sexp ' (w) = 0 then it follows that for real k>0 : sexp ' (w+k) = 0. Proof : chain rule exp^[k] is analytic : sexp(w+k) = exp^[k](sexp(w)) sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0 Hence we get a contradiction : sexp is not nonpolynomial analytic near w (or w + k). Conclusion there is no w such that sexp'(w) = 0. Consequences : since 0 < sexp ' (z) < oo slog ' (z) is also 0 < slog ' (z) < oo since exp^[k](v) = sexp(slog(v)+k) D exp^[k](v) = sexp ' (slog(v)+k) * slog ' (v) = nonzero * nonzero = nonzero.. => 0 < exp^[k] ' (z) < oo Tommy's theorem Strongly related to the TPID 4 thread and some recent conjectures of sheldon. the analogue difference is not understood yet. (posted that already) regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread [2014] tommy's theorem sexp ' (z) =/= 0 ? - by tommy1729 - 06/17/2014, 12:18 PM RE: [2014] tommy's theorem sexp ' (z) =/= 0 ? - by sheldonison - 06/17/2014, 01:25 PM

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